次の4つの不定積分を求めます。 (1) $\int \tan^4 x dx$ (2) $\int \frac{dx}{\sin 2x}$ (3) $\int \frac{1}{1-\sin x} dx$ (4) $\int (\sin^3 x - \cos^3 x) dx$

1. 問題の内容

次の4つの不定積分を求めます。

(1) $\int \tan^4 x dx$

(2) $\int \frac{dx}{\sin 2x}$

(3) $\int \frac{1}{1-\sin x} dx$

(4) $\int (\sin^3 x - \cos^3 x) dx$

2. 解き方の手順

(1) $\int \tan^4 x dx$ の場合：

$\tan^4 x = \tan^2 x \cdot \tan^2 x = \tan^2 x (\sec^2 x - 1) = \tan^2 x \sec^2 x - \tan^2 x = \tan^2 x \sec^2 x - (\sec^2 x - 1)$

$\int \tan^4 x dx = \int (\tan^2 x \sec^2 x - \sec^2 x + 1) dx = \int \tan^2 x \sec^2 x dx - \int \sec^2 x dx + \int dx$

ここで、$\int \tan^2 x \sec^2 x dx = \int (\tan x)^2 (\tan x)' dx = \frac{1}{3} \tan^3 x + C_1$

$\int \sec^2 x dx = \tan x + C_2$

$\int dx = x + C_3$

よって、$\int \tan^4 x dx = \frac{1}{3} \tan^3 x - \tan x + x + C$

(2) $\int \frac{dx}{\sin 2x}$ の場合：

$\sin 2x = 2 \sin x \cos x$ より、$\int \frac{dx}{\sin 2x} = \int \frac{dx}{2 \sin x \cos x} = \frac{1}{2} \int \frac{dx}{\sin x \cos x} = \frac{1}{2} \int \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} dx = \frac{1}{2} \int (\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}) dx = \frac{1}{2} \int (\tan x + \cot x) dx = \frac{1}{2} \int (\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}) dx = \frac{1}{2} (-\ln |\cos x| + \ln |\sin x|) + C = \frac{1}{2} \ln |\frac{\sin x}{\cos x}| + C = \frac{1}{2} \ln |\tan x| + C$

別の解法：$\int \frac{dx}{\sin 2x} = \int \frac{1}{2 \sin x \cos x} dx = \int \frac{1}{2 \tan x \cos^2 x} dx = \frac{1}{2} \int \frac{\sec^2 x}{\tan x} dx$。 $u = \tan x$ とすると、$du = \sec^2 x dx$ より、$\frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |\tan x| + C$

(3) $\int \frac{1}{1-\sin x} dx$ の場合：

$\int \frac{1}{1-\sin x} dx = \int \frac{1+\sin x}{(1-\sin x)(1+\sin x)} dx = \int \frac{1+\sin x}{1-\sin^2 x} dx = \int \frac{1+\sin x}{\cos^2 x} dx = \int (\frac{1}{\cos^2 x} + \frac{\sin x}{\cos^2 x}) dx = \int \sec^2 x dx + \int \frac{\sin x}{\cos^2 x} dx = \tan x + \int \frac{\sin x}{\cos^2 x} dx$

$\int \frac{\sin x}{\cos^2 x} dx$ について、$u = \cos x$ とすると、$du = -\sin x dx$ より、$\int \frac{\sin x}{\cos^2 x} dx = \int \frac{-du}{u^2} = \frac{1}{u} + C = \frac{1}{\cos x} + C = \sec x + C$

よって、$\int \frac{1}{1-\sin x} dx = \tan x + \sec x + C$

(4) $\int (\sin^3 x - \cos^3 x) dx$ の場合：

$\sin^3 x = \sin x (1-\cos^2 x) = \sin x - \sin x \cos^2 x$

$\cos^3 x = \cos x (1-\sin^2 x) = \cos x - \cos x \sin^2 x$

$\int (\sin^3 x - \cos^3 x) dx = \int (\sin x - \sin x \cos^2 x - \cos x + \cos x \sin^2 x) dx = \int \sin x dx - \int \cos x dx - \int \sin x \cos^2 x dx + \int \cos x \sin^2 x dx = -\cos x - \sin x - \int \sin x \cos^2 x dx + \int \cos x \sin^2 x dx$

$\int \sin x \cos^2 x dx$ について、$u = \cos x$ とすると、$du = -\sin x dx$ より、$\int \sin x \cos^2 x dx = \int -u^2 du = -\frac{1}{3} u^3 + C = -\frac{1}{3} \cos^3 x + C$

$\int \cos x \sin^2 x dx$ について、$u = \sin x$ とすると、$du = \cos x dx$ より、$\int \cos x \sin^2 x dx = \int u^2 du = \frac{1}{3} u^3 + C = \frac{1}{3} \sin^3 x + C$

よって、$\int (\sin^3 x - \cos^3 x) dx = -\cos x - \sin x - (-\frac{1}{3} \cos^3 x) + \frac{1}{3} \sin^3 x + C = -\cos x - \sin x + \frac{1}{3} \cos^3 x + \frac{1}{3} \sin^3 x + C$

3. 最終的な答え

(1) $\int \tan^4 x dx = \frac{1}{3} \tan^3 x - \tan x + x + C$

(2) $\int \frac{dx}{\sin 2x} = \frac{1}{2} \ln |\tan x| + C$

(3) $\int \frac{1}{1-\sin x} dx = \tan x + \sec x + C$

(4) $\int (\sin^3 x - \cos^3 x) dx = -\cos x - \sin x + \frac{1}{3} \cos^3 x + \frac{1}{3} \sin^3 x + C$